I am trying to factorize the expression: $$(x+y+z)^p-[(-x+y+z)^p+(x-y+z)^p+(x+y-z)^p]$$ where $p$ is an odd prime and $x,y,z$ are any non-zero integers.
I know that it is divisible by $pxyz$.
How do I find/write the remaining factor(s) in summation notation? I tried to use the trinomial expansion but I keep getting lost. Any hints?
On
By using multinomial theorem, it's easy to see that after cancellation, all terms in your expression will have coefficient divisible by $p$ (since only terms which don't have such coefficient when expanding, say, $(x+y-z)^p$ are $x^p,y^p,-z^p$, and these cancel out from all the sums). Now we need to show all terms not containing all of $xyz$ get cancelled out (as then all leftover terms will be of the form $kpx^ay^bz^c,k\in\Bbb Z,a,b,c>0$ which will end the proof). Suppose the term doesn't contain $z$, so is of the form $x^ay^{p-a}$. We have from binomial theorem $(x+y+z)^p=\text{(stuff containing $x$)}+(y+z)^p=\binom{a}{p}y^az^{p-a}+\text{(other terms)}$. It's easy to see that from $(-x+y+z)^p,(x-y+z)^p,(x+y-z)^p$ you will get, in some order, $\binom{a}{p}y^az^{p-a}+\binom{a}{p}y^az^{p-a}-\binom{a}{p}y^az^{p-a}+\text{(other terms)}$. So, all in all, terms containing $y^az^{p-a}$ will all cancel out and we are home.
NOTE: I would write this as $A = -x+y+z,$ $B = x-y+z,$ $C=x+y-z,$ thus demanding that $A,B,C$ are all odd or all even. Then the expression is $$ (A+B+C)^p -A^p - B^p - C^p. $$ Probably good for something
Appears that I should have written $p=5$ as $$ 40xyz(2x^2 + 2 y^2 + 2 z^2) $$ This way, we get a consistent first term, $$ 8pxyz \left( \; \frac{p-1}{2} \, x^{p-3} + \mbox{other} \right) $$
bit of a mess to typeset. I did primes $3,5,7,11$ in gp-pari, wrote out the factors, each in a recognizable pattern. Don't see anything overall except for the even exponents and considerable symmetry.
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